Prof. Ramesh Bansal
Dept. of Electrical Engineering
University of Sharjah
rbansal@sharjah.ac.ae; M9-210
Learning objectives
After completing this chapter, you should be able to learn
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Ideal Transformer With An Imperfect Core
Ideal Transformer With Loose Coupling
Equivalent Circuit of a Practical Transformer
Transformer Taps
Losses and Transformer Rating
Simplifying the Equivalent Circuit
Voltage Regulation
Measuring Transformer Impedance-OC & SC tests
Per Unit Method and Per Unit Impedance of a Transformer
 Imperfect core means the permeability is not infinity. And it
has electric losses due to eddy currents and hysteresis loops.
In order to get electric model of the magnetic core, a
resistance Rm is used to represent iron losses in the
transformer. Also, magnetizing reactance Xm is used in shunt
with the resistance to hold the magnetizing current and
represent main flux in the circuit.
The equivalent circuit for
the real iron core is shown
 The total current taken by the real core,
(imperfect), has two different components; one
component that generates the losses If and the
other component that develops the field in the
transformer Im. The sum of these two
components are called exciting current I0. (upto
5 %) of the full-load, (maximum current).
 Example 10.1: A large transformer operating at no-load
draws an exciting current Io of 5 A when the primary is
connected to a 120 V, 60 Hz source, (circuit shown
previously). From a wattmeter test it is known that iron
losses are equal to 180 W. Calculate
A. The reactive power absorbed by the core
B. The value of Rm and Xm
Ans:572 Var, 80 Ohm, 25.2
C. The value of If, Im and Io
Ohm 1.5A, 4.8A, 5A
 Solution: Io = 5 A , E1=120 V, F=60 Hz,. P0=180 W.
A. S=I0E1=5x120=600VA

=P+jQ=180+jQ
Q=√(6002-1802)=572 var
B. 𝑅𝑚 = E12/P0=1202/180=80 ohm,
𝑋𝑚 = E12/Q=1202/572=25.2 ohm
C. Io =5A,
If =E1/Rm=1.5A
Im =E1/Xm=4.8A
 It is mentioned before, the core is not perfect, which
means that the iron core does not collect all the flux lines
and the flux lines will be distributed between the air and
core. The flux distribution in the domain is shown in the
figure below.
The mutual flux will develop Ep and Es in both windings, (shown in
the figure of previous slide).
E p  4.44 fN1m
Es  4.44 fN 2m
 The leakage flux,  f 1 and  f 2 in both sides interact with the windings
and develop voltages as,
E f 1  4.44 fN1 f 1
E f 2  4.44 fN 2 f 2
The simplest way to represent these four voltages
is to arrange the winding as shown below.
The primary and secondary
voltages Ef1 and Ef2 are the
voltage drop across reactance,
which are called primary and
secondary leakage reactance.
These reactances can be
calculated as,
X f1 
Ef1
I1
,Xf2 
Ef 2
I2
Example 10.2: The secondary winding of a transformer
has 180 turns. When the transformer is under load, the
secondary current has an effective value of 18 A, 60 Hz.
Furthermore, the mutual Фm has a peak value of 20 mWb.
The secondary leakage flux Фf2 has a peak value of 3
mWb. Calculate
a. The voltage induced in the secondary winding by its
leakage flux
b. The value of the secondary leakage reactance
c. The value of E2 induced by the mutual flux Фm
Ans:143.9 V, j8Ω, 959
 The circuit shown below indicates exact equivalent circuit of the
transformer.
 Each winding is represented by an internal resistance and a
leakage reactance. The two voltages E1 and E2 result from interaction
between the mutual flux and the two windings.
 The electrical equivalent model of the iron core is given as parallel
branch of Rm and Xm.
Ideal Transformer
 Due to the voltage drops in transmission system lines,
the voltage in a particular region of a distribution system
may consistently be lower than normal. Under these
conditions the voltage across the secondary is less than
the normal. To correct this problem taps are provided either
on the primary or the secondary sides to correct the
voltage in the system.
 Some transformers are designed to change the taps
automatically whenever the secondary voltage is not close
to the preset value.
 Like any other machines, a transformer has losses, they are
composed of the following:
1- I2R losses in the windings
2- Hysteresis and eddy losses in the core
3- Stray losses due to current induced in the tank and metal support by
the primary and secondary leakage flux
 The losses appear in the form of heat and produce
- an increase in the temperature
- a drop in the efficiency
 The Apparent power rating of the transformer is the multiplication
of the nominal, (maximum allowable for safe operation), voltage and
nominal current. It is expressed in VA or kVA or MVA.
 The rated VA, frequency, nominal voltage are always shown on the
nameplate.
Example 10.4
The nameplate of a distribution transformer indicates 250
kVA, 60 Hz, primary 4160 V, secondary 480 V.
a. Calculate the nominal primary and secondary currents
b. If we apply 2000 V to the 4160 V primary, can we still
draw 250 kVA from the transformer. (for safe operation)
Ans:60 A, 521 A, S = 120 kVA
 The exact circuit of the transformer gives more details than what
are needed for the analysis of the transformer. Two different cases will
be studied for the transformer analysis.
1- At No Load
At no-load condition, the current in the secondary side is zero. The
only current which is considered is the no-load current, (exciting
current). Therefore, the drop over the series impedance at the primary
side is almost negligible (IO is so small). Thus, the parallel branches
can be moved from its location to be close to the supply and simplify
the circuit as shown below.
2- At Any Load
 When the transformer is fully loaded, (rated current), or with any
load. The current in the primary side is about 20 times the no-load
current. Therefore, the no-load current can be neglected with respect
to the load current in the primary side. The approximate equivalent
circuit is shown below.
 The approximate circuit can be referred to either primary side or
secondary side.
The circuit referred
primary side.
to
 Voltage regulation (VR) of transformer can be
calculated on primary or secondary side
 Let the primary voltage held constant at its rated value.
The voltage regulation on secondary side is defined as,
VR 
Enl  E f l
E fl
 The voltage regulation depends on several factors
 Power factor
 Load impedance
 Load current
 The previous equation can be re-written into the
following equation:
VR 
Es  nl  Es  f l
Es  fl
Example 10.5: A single-phase transformer rated at 3000
kVA, 69 kV/4.16 kV, 60 Hz has a total internal impedance
Zp of 127 Ω, referred to primary side (resistive part is
neglected).
Calculate:
a. The rated primary and secondary currents
b. The voltage regulation from no-load to full-load for a
2000 kW resistive load, knowing that the primary supply
voltage is fixed at 69 kV.
c. The primary and secondary currents if the secondary
is accidentally short circuit.
 Two important tests can be applied on the transformer
to determine the impedances of the shunt branches and of
the series branch. The first test is called open-circuit test
and the other test is called short-circuit test.
Open-Circuit Test
 In this test, the transformer secondary side is opencircuit. Voltmeter, ammeter and wattmeter are installed
and connected at the primary side as shown in the figure
at right side.
 The equivalent circuit for open-circuit test is similar to
the normal open circuit equivalent for the transformer.
This circuit is depicted at the right side.
 The readings of voltmeter, ammeter and wattmeter
are used to get the value of Xm and Rm.
 The analysis of the circuit depicted in the next slide,
the main idea is that the transformer can be dealt as if it is
a circuit with two parallel branches.
Analysis of Open-Circuit
S  EP I o
Q 
S 2  P2
Rm
E 2P

Pm
Xm
E 2P

Qm
EP
a 
Es
Short-Circuit Test
 In this test, the secondary side is short-circuit and the
applied voltage is slightly increased till the primary current
is equal to or close to the primary rated current. (The
applied voltage is approximately 5% of the rated primary
voltage).
 The voltage, current, and power are measured. From
these measurements, the series parameters can be
calculated as follows:
Analysis of Short-Circuit
Z p  Z1eq
E sc
Psc

, R p  R1eq  2
I sc
I sc
X p  X 1eq 
Z p2  R p2
Example 10.6: during a short circuit test on a transformer rated 500
kVA, 69 kV/4.16 kV, 60 Hz, the following voltage, current and
power measurements were made. Terminals X1, X2 were in short
circuit (see figure)
Esc=2600, Isc=4A, Psc=2400W.
Calculate the value of the reactance and resistance of the transformer
referred to the HV side.
(Ans.: Rp=150 ohm, Xp=632 Ohm)
Example 10.7: An open-circuit test was conducted on the transformer
given in example 10-6. (transformer rated 500 kVA, 69 kV/ 4.16 kV, 60
Hz). The following results were obtained when the low-voltage winding
was excited (in some cases, such as in a repair shop, a 69 kV voltage
may not be available and the open-circuit test has to be done by
exciting the LV winding at its rated voltage).
Es=4160 V
Io=2 A
Pm=5000 W
calculate:
a. The values of Xm and Rm on the primary side, (referred)
b. The efficiency of the transformer when it supplies a load of 250
kVA, whose power factor is 80% (lagging).(giving that Rp=150 Ω)
Solution
Ans: 96.6%
 The per unit method is a way that refer any value to its
corresponding base value. For the impedance, the
impedance in ohm should be divided over a based
impedance to get its per unit value. For the transformer,
the series and shunt impedance of the transformer is
commonly expressed in per unit in the name-plate.
Z base
Z pu
Vbase

I base
Z [ ]

Z base []
 For transformer:
Z base  p  Z base 1
Z pu 1 
Z p [ ]
Z base 1[ ]
Z base  s  Z base  2
Z pu  2
Vrated  p
Vbase 1
Vrated 1



I base 1
I rated 1
I rated  p

Z1eq []
Z base 1[ ]
Vbase  2
Vrated  2
Vrated  s



I base  2
I rated  2
I rated  s
Z 2 eq []
Z s [ ]


Z base  2 []
Z base 2 [ ]
Example 10.8
A transformer rated 250 kVA, 4160 V/480 V, 60 Hz has an impedance
of 5.1%. Calculate
a. The base impedance on the primary and secondary side
b. The total internal impedance Zp of the transformer referred to the
primary side
Problem 10.11 to 10.31